integral of (2x+3)/(5+4x+4x^2)

解

\int \frac{2 x + 3}{5 + 4 x + 4 x ^{2}} d x

\frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2}) + C

解答ステップ

\int \frac{2 x + 3}{5 + 4 x + 4 x ^{2}} d x

拡張

\frac{2 x + 3}{5 + 4 x + 4 x ^{2}} : \frac{2 x}{5 + 4 x + 4 x ^{2}} + \frac{3}{5 + 4 x + 4 x ^{2}}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{5 + 4 x + 4 x ^{2}} d x + \int \frac{3}{5 + 4 x + 4 x ^{2}} d x

\int \frac{2 x}{5 + 4 x + 4 x ^{2}} d x = \frac{1}{4} ln 4 x^{2} + 4 x + 5 - \frac{1}{4} arctan (x + \frac{1}{2})

\int \frac{3}{5 + 4 x + 4 x ^{2}} d x = \frac{3}{4} arctan (x + \frac{1}{2})

= \frac{1}{4} ln 4 x^{2} + 4 x + 5 - \frac{1}{4} arctan (x + \frac{1}{2}) + \frac{3}{4} arctan (x + \frac{1}{2})

類似した元を足す:

- \frac{1}{4} arctan (x + \frac{1}{2}) + \frac{3}{4} arctan (x + \frac{1}{2}) = \frac{1}{2} arctan (x + \frac{1}{2})

= \frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2})

定数を解答に追加する

= \frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2}) + C